Talk:Brouwer fixed-point theorem/Archive 1

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Archive 1

Proofs

Why is wikipedia not the place to reroduce a long proof? —Preceding unsigned comment added by 130.238.5.5 (talk • contribs) 22 September 2005

What? El_C 20:10, 22 September 2005 (UTC)

See Wikipedia:WikiProject Mathematics/Proofs and the math style manual. Oleg Alexandrov 22:44, 22 September 2005 (UTC)

Non-constructive

The history section seems to imply that the theorem does not have a constructive proof. Is there a way to formalize (and prove) this statement? AxelBoldt 05:37, 9 June 2006 (UTC)

Ok, I found in [1] that there are in fact algorithms to approximate a fixed point. AxelBoldt 03:34, 12 June 2006 (UTC)

Outline of Proof

I believe that there's an error outline of the proof given - in particular the induced transformation from the disk to its boundary need not be a retraction (the points on the boundary need not be fixed by the transformation). At the same time this is only a small error in that it's true that there's no continuous mapping from the disk to its boundary (retraction or not).

I'm not sure what you mean. The map described is the one that sends x to the point on the boundary given by following the directed line going from f(x) to x. So if x is on the boundary, it gets fixed. --C S (Talk) 03:36, 1 May 2006 (UTC)

Sorry, you're completely right, I thought the ray was in the other direction.

However there is still a problem, why does

F(x)

first have to be continuous (the above section shows that it is, but this should be in the outline). The biggest problem I see is that

F:D^{2}\to S^{1}

is not necessarily surjective! --Dark Side of the Moon 16:59, 28 August 2006 (UTC)

F is necessarily surjective, as it maps every element of

S^{1}

to itself. As for continuity, this follows from the explicit formula for F(x) that someone has given above. --Zundark 17:57, 28 August 2006 (UTC)

I was mistaken; I had misread and thought that the point F(x) was created by moving from x towards f(x) which would cause problems. The other way, the way it is in the article, will infact work. --Dark Side of the Moon 21:10, 28 August 2006 (UTC)

Induced map goes the wrong way

Since the retraction maps D² to S¹, its induced map goes from π₁(D²) to π₁(S¹) rather than the other way around. The injective map from π₁(S¹) to π₁(D²) is induced by the inclusion, while the retraction induces a surjective map in the other direction. 69.234.55.7 20:45, 15 November 2007 (UTC)

Minor point

It seems to me that in the "Notes" near the beginning, where it mentions that the fixed point theorem also applies to objects homeomorphic to the closed unit n-disc, the property "bounded" should be removed from the parentheses. Of course that would be true if the homeomorphism is into some R^m, by the Heine-Borel Theorem. But the homeomorphism need not be to some R^m; it needn't even be to a metric space. Maybe in parentheses, instead of saying "closed, bounded," it should just say "compact." Kier07 22:56, 12 December 2006 (UTC)

The homeomorphic space might not be metric, but it is compact metrizable, and in a compact metrizable space, every metric is bounded. —Preceding unsigned comment added by 66.191.126.159 (talk) 06:36, 21 November 2007 (UTC)

General editing

Some of the HTML was weird, and I went through and fixed these while reducing some of the spacing in superscripts. A few sentences were written in a way which could have been a little more encyclopaedic, and I removed some plural third-person references where I was able to. There could still be a little bit of a tidy-up on this article, and I certainly might not have done all of the edits correctly since there were a number of them. Xantharius (talk) 17:29, 17 April 2008 (UTC)

Hirsch's proof

The claim that leads to the contradiction is not true as stated: it can happen that the inverse image of a regular value of F is a topological circle in Dⁿ with one point on the boundary sphere Sⁿ⁻¹. This is compatible with the map F being a retraction. What then? Arcfrk (talk) 09:49, 23 March 2008 (UTC)

um, are you trying to cause trouble? :-) Anyway, I fixed it up a bit to avoid your objection. --C S (talk) 18:02, 27 July 2008 (UTC)

Another proof in the case n=2

"The case n = 2 can also be proven by contradiction based on a theorem about non-vanishing vector fields." Why not write the name of the theorem, or at least give a link to it? --SuneJ (talk) 15:32, 5 August 2008 (UTC)

One of the statements in the "Intuitive proof" contains an error

I have removed the statement from the intuitive proof that claims that the indicated zero set has to contain a line. It does not -- it is an easy exercise, for instance, to construct functions of the sort indicated whose zero sets are things like the "topologist's sine curve", which definitely does not contain a line!

Making the "intuitive proof" precise is actually quite difficult. One would need to know something about dimension theory, but the usual proofs of the basic facts in dimension theory needed to prove the needed results actually use Brouwer's fixed point theorem! Someone should write a warning in the beginning of it to this effect, but I don't have enough free time to do so. —Preceding unsigned comment added by 24.41.92.61 (talk) 04:27, 6 July 2009 (UTC)

Illustration

Since we require a resolution that is high of the graphpaper why bother saying its a graph paper, just makes things confusing?

Assume that you have four squares on a paper i.e. one centerpoint, now it is easy to place the paper so that it is not directly above a point? —Preceding unsigned comment added by 85.24.185.96 (talk) 18:04, 4 September 2009 (UTC)

Image problem

File:Brouwer_fixed_point_theorem_retraction.svg isn't displaying properly for me. Sławomir Biały (talk) 00:05, 19 November 2009 (UTC)

For me it's fine both with Firefox 3.5.5 and with Internet Explorer 8. Such problems can be highly browser dependent and are very hard to debug. Hans Adler 00:11, 19 November 2009 (UTC)

I'm having problems in Konqueror 3.5.5 and Iceweasel 2.0.0.8. The svg file also doesn't render properly in inkscape. Sławomir Biały (talk) 02:44, 19 November 2009 (UTC)

Also doesn't work in Safari 4.0.3 and Firefox 3.0.15 (on a MAC running OSX 10.3). Sławomir Biały (talk) 02:47, 19 November 2009 (UTC)

Maybe its a locale issue? The image was uploaded on de: first. Sławomir Biały (talk) 02:59, 19 November 2009 (UTC)

Same problem in Google Chrome. (FYI: the problem I am seeing is that the parentheses appear before F and f, rather than around the function argument.) Sławomir Biały (talk) 18:37, 19 November 2009 (UTC)

Ah, since you didn't say what it was I thought it was something more obvious. I had the same problem with my work computer, which was getting a generated PNG file rather than the original SVG. The SVG rendered fine on my platform. I think I have fixed it. Hans Adler 19:01, 19 November 2009 (UTC)

Fabulous. That did the trick. :-) Sławomir Biały (talk) 02:57, 20 November 2009 (UTC)

non-constructive, again

"While Brouwer preferred constructive proofs, ironically, the original proofs of his great topological theorems were not constructive, and it took until 1967 for constructive proofs to be found."

Well, in fact there is no constructive proof of the fixed-point theorem: it is false (not realisable)! And, as I understand Brouwer eventually knew a refutation of it, and renounced his own theorem. What we have proofs of is constructive substitutes: there is a fixed point up to epsilon, but this does not always give an exact fixed point because the approximations do not always converge as epsilon goes to zero. --99.245.206.188 (talk) 03:32, 21 January 2010 (UTC) (PS: why does an article on Brouwer's fixed point theorem start with a picture of Poincaré?)

Generalization for compact connected orders

Under "Generalizations" it is stated that if $L$ is a compact, connected order topology, then any continuous function from $L^{n}$ to itself has a fixed point, but there is no reference given. Does anyone have a proof for this? -- AndreasF82 (talk) 12:57, 20 October 2010 (UTC)

Notes

Correct me if i'm wrong, but I think that there is an error in the counterexample for domains which aren't closed. You claim that for the unit disc, which is not closed: $f(x,y)=\textstyle ({\frac {1}{2}}(x+{\sqrt {1-y^{2}}}),y)$ is a function with no fixed point. But if we set $x={\frac {1}{2}}$ and $y={\sqrt {\frac {3}{4}}}$ , then $f(x,y)=(x,y)$ .--84.73.188.172 (talk) 18:07, 19 January 2011 (UTC)

Congratulations! You found an error that has been in the article since June 2006!

The idea is valid, though. I will replace the example with a hopefully correct one using the homeomorphism of the interval (-π/2,π/2) with the real line via the tan and arctan functions. Since I have just made this up and did not put much thought into it, it would be nice if you could check it. Hans Adler 20:46, 19 January 2011 (UTC)

Yes, I guess the corrected version is ok. Another, maybe simpler example, would be the open interval (0,1) and the function

f(x)={\frac {x}{2}}

. Since

the function doesn't have to be bijective, this should also work out. --84.73.188.172 (talk) 14:28, 22 January 2011 (UTC)

Illustrations: Ordinary map of a country

"Similarly: Take an ordinary map of a country, and suppose that that map is laid out on a table inside that country. There will always be a "You are Here" point on the map which represents that same point in the country." A country is not necessarily convex (can you name one that is?), which is required in the Brouwer's fixed-point theorem, so I don't see how this is an illustration of the theorem. An ordinary map is usually a contraction mapping, so this can be proved using the Banach fixed-point theorem instead. --82.130.37.20 (talk) 18:07, 7 February 2012 (UTC)

On the other hand, if the table is convex (or more generally, if the map is inside a convex subset of the country), the Brouwer's fixed-point theorem can be used to prove that there is a fixed point in the part of the map that represents the table (or the convex subset). But this is not as simple as the example currently given in the article. --82.130.37.20 (talk) 18:24, 7 February 2012 (UTC)

File:Poincare.jpg Nominated for Deletion

Please do remove the picture. Who expects to see a photograph of Poincare on a page named after Brouwer!

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The Liquid in a glass example

should either be clarified or deleted. It certainly isn't true as stated. I suppose that whoever wrote it was thinking of continuum fluid dynamics but real fluids aren't continua, they're made of atoms all of which undergo thermal motion. —Preceding unsigned comment added by 65.19.15.202 (talk) 15:43, 24 April 2009 (UTC)

Well that is about how literal you want to be in describing "the liquid" in general as a "way of understanding" it is correct - a point need not be an atom in a mathematical view of the liquid. Although maybe not the exact physical interpretation. Honestly i really don't see the problem here. Gillis (talk) 21:57, 24 April 2009 (UTC)

A point need not be an atom in a mathematical view of the liquid? Although maybe not the exact physical interpretation? This is what I would call a clarification, which is exactly what the OP requested. That's the problem here. It's a very small problem, certainly. And most readers can probably understand the intended concept the way it is stated. Perhaps the word cocktail could be prefaced with (mathematically idealized) as a way to head off any argument about discrete molecules. I really believe such a minor mod would reduce the frustration some readers feel when we encounter apparent disconnects from reality such as this. Thanks for your time.--Twixter (talk) 00:26, 13 July 2012 (UTC)

Hex in multiple dimensions?

The article says:

A quite different proof can be given based on the game of Hex. The basic theorem about Hex is that no game can end in a draw. This is equivalent to the Brouwer fixed point theorem for dimension 2. By considering n-dimensional versions of Hex, one can prove that in general that Brouwer's theorem is equivalent to the "no draw" theorem for Hex.

But how does one play an "n-dimensional version of Hex"?

Consider the original 2D hexboard as being made from a lattice, where one connects the lower left corner of a square of a lattice to the upper right corner with an edge. These added diagonals make it so you can have 6 neighbors instead of 4 (on the original lattice). It's easy to see how to cut out an n by n Hex board from this modified lattice. In general, to create an n x n .... x n board, consider a lattice in R^m (where m is the number of dimensions of the board) and then in each m-dimensional cube add a diagonal. Then cut out a board as in 2D.

Each player has an opposite pair of sides as before, but now some sides belong to neither player. The game is played the same way as before. It doesn't appear to be so interesting to play, but people have devised other higher dimensional versions which are probably more fun and interesting mathematically. In any case, in the version I described, there can never be a draw, and this no-draw result is equivalent to the Brouwer fixed point theorem. --C S (Talk) 07:03, 15 January 2006 (UTC)

Does the number of players remain 2? What exactly is the object on these higher dimensional grids? For example, in 3 dimensions, a "side" refers to a 2-dimensional face of the lattice, yes? If the object for both sides is to connect any point on one of your border faces with a point on the opposite border face via a continuous path, then the blocking nature of the game disappears, since both sides can achieve their object without having to block the other. If the lattice were filled with stones, most likely there would be multiple winning paths for both sides, which would be an illegal position, inasmuch as the game ends the moment either side achieves the objective. It may well be the case that a draw is impossible, but is this truly all you need to prove BFPT?

Maybe the object on a 3D grid needs to be as follows: One player need only construct a path connecting his opposite faces as described, but the other player must block all possible paths by the opponent, by building a membrane which stretches through the grid, connecting an entire path which crosses one of his border faces to a path across his other border face, with no holes. Of course this is an unfair game, but now it is impossible for both sides to achieve a winning path on a filled grid.--Twixter (talk) 00:56, 13 July 2012 (UTC)

elemantary proof with stokes' theorem

first, the retraction is given by:

File:Theorem of brouwer-F.png

Illustration of the retraction

F(x):=x+\left({\sqrt {1-|x|^{2}+\left\langle x,{\frac {x-f(x)}{|x-f(x)|}}\right\rangle ^{2}}}-\left\langle x,{\frac {x-f(x)}{|x-f(x)|}}\right\rangle \right){\frac {x-f(x)}{|x-f(x)|}}

and it can be proved using the strokes' theorem for differential forms.

for $\omega ^{n-1}:=F^{1}\,dF^{2}\wedge \cdots \wedge dF^{n}$ , $d\omega ^{n-1}=0$ , so you get: $0=\int _{D^{n}}\mathrm {d} \omega ^{n-1}=\int _{S^{n-1}}\omega ^{n-1}=\int _{S^{n-1}}x_{1}dx^{2}\wedge \cdots \wedge dx^{n}=vol(D^{n})\neq 0$ . first = because the jacobian is 0 by theorem of implicit functions.

See the german page as example. this could be integrated. ~ibotty

Looks a bit messy to me.... Oleg Alexandrov (talk) 16:24, 9 February 2006 (UTC)

It's not that messy! Don't be scared by the symbols, Oleg :-) I think with a little reworking it would make a fine addition. This is a pretty famous proof. Unfortunately, I take a bit of an issue with "elementary" describing this proof. The proof, as given, only proves Brouwer's theorem for sufficiently smooth maps f, since one needs to take the exterior derivative. Luckily, we can homotope f to be smooth while still keeping the map fixed point free (we can pick a straight-line homotopy that moves every point less than epsilon, where epsilon is smaller than the minimal distance between x and f(x)); however, this is not so trivial to show and can be regarded as a technicality that makes the entire proof not as elementary. And of course, we could debate, if we wished, whether this whole business of Stoke's theorem and smoothing maps is really more elementary than some simple homology (or homotopy) theory. Personally, I think the only kind of proof of Brouwer that really qualifies as elementary are the ones involving some form of coloring trick, e.g. Sperner's lemma or Hex. --C S (Talk) 01:46, 2 April 2006 (UTC)

Hmmm...actually I see that since I last closely perused the article, somebody has added another proof of Brouwer for smooth maps. Per my comment right above, this is actually a proof for continuous maps also; I'll add that to the article. --C S (Talk) 10:42, 2 April 2006 (UTC)

There has been some little discussion on math.stackexchange here about the validity of this proof. --Sean Eberhard (talk) 09:55, 12 September 2012 (UTC)

Fascinating discussion with some great insights. Nonetheless, there is an "obvious" way, at least to a topologist. Since we know the smooth maps have a fixed point and the continuous map can be approximated by smooth maps, just show the set of continuous maps with a fixed point is closed. Not hard. --C S (talk) 22:43, 8 May 2013 (UTC)

Stokes theorem may well be derived from the Brouwer Fixed Point Theorem, so I would be wary of using it. But I am not certian of this, as I have not read the proof. --Dark Side of the Moon 17:02, 28 August 2006 (UTC)

It's not. —Preceding unsigned comment added by 66.191.126.159 (talk) 06:20, 21 November 2007 (UTC)

Is convexity required?

Under "Statement", it is written that the theorem holds only for a convex set. However, under "Notes", it is written that: "For the same reason it holds for every set that is homeomorphic to a closed ball (and therefore also closed, bounded, connected, without holes, etc.).". As far as I know, there are non-convex sets which are homeomorphic to balls. So, does the theorem holds for non-convex set or not?

It could be good to show an example of a non-convex set on which the result does not hold. --Erel Segal (talk) 06:31, 19 December 2014 (UTC)

For a compact convex set, all of which are homeomorphic to a ball. So e.g. an annulus is compact non-convex, and a rotation maps it to itself without fixpoint. --206.214.242.230 (talk) 18:30, 23 January 2015 (UTC)

A proof using the oriented area

"Differentiating under the sign of integral it is not difficult to check that φ′(t)=0 for all t" — really? How to check it? Maybe the Milnor-Rogers-Gröger approach is meant? There, φ is polynomial (not just smooth) and appears to be constant near 0, therefore everywhere. Boris Tsirelson (talk) 08:51, 15 September 2015 (UTC)

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"Constructive proof"?

The article sais "The first algorithm to construct a fixed point was proposed by H. Scarf." and also "Kellogg, Li, and Yorke turned Hirsch's proof into a constructive proof by observing that..."
I'm wondering if it's indeed a constructive proof, since Brower's theorem for one dimension is equivalent to intermediate value theorem, which does not admit a constructive proof.
See for example this discussion in MathOverflow. Nachi (talk) 16:58, 4 February 2018 (UTC)

This is a symptom of different people using "constructive" to mean different things. The paper by Kellog, Li, and York really is titled "A Constructive Proof of the Brouwer Fixed-Point Theorem and Computational Results". But they are working in numerical analysis, not in constructive mathematics. So perhaps all that they mean by 'constructive proof' is that their proof can be used to obtain a numerical algorithm to approximate a fixed point. I am not completely sure what they mean by constructive, though, as I look at their paper. They also assume that the map is not only continuous, but twice differentiable. In the sense of many branches of constructive mathematics, it is known that the fixed point theorem implies nonconstructive principles such as LLPO, and so the fixed point theorem is not constructive in the sense of those branches. — Carl (CBM · talk) 17:25, 4 February 2018 (UTC)

Thank you for the answer. It makes it clear.

Unless I completely not aware of the usual use of "constructive" in mathematics, I guess the best way to describe KLY version of Hirsch's proof is simply write "numerical algorithm" or "computable method", instead of "constructive". Also in the description of Scarf's proof the word "construct" should be replaced by "calculate". Nachi (talk) 20:25, 4 February 2018 (UTC)